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jonnymiller
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Hi all, I am still working on second derivative tests having trouble with finding the critical points.
The problem is find the critical points of f(x,y)=sinx+siny+cos(x+y) for x between 0 and pie/4 and y between 0 and pie/4. Classify each as a local min, max, or saddle point.
f(x,y)=sinx+siny+cos(x+y)
I found the partials to be:
derivative of f with respect to x = fx=cosxsin(x+y)
derivative of f with respect to y = fy=cosysin(x+y)
I am having trouble solving the partials for 0.
 one year ago
 one year ago
jonnymiller Group Title
Hi all, I am still working on second derivative tests having trouble with finding the critical points. The problem is find the critical points of f(x,y)=sinx+siny+cos(x+y) for x between 0 and pie/4 and y between 0 and pie/4. Classify each as a local min, max, or saddle point. f(x,y)=sinx+siny+cos(x+y) I found the partials to be: derivative of f with respect to x = fx=cosxsin(x+y) derivative of f with respect to y = fy=cosysin(x+y) I am having trouble solving the partials for 0.
 one year ago
 one year ago

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jonnymiller Group TitleBest ResponseYou've already chosen the best response.0
For fx=cosxsin(x+y)=0 fy=cosysin(x+y)=0 x=pie/6 y=pie/6 I did this by trial and error. Does anyone know how to solve it algebraically?
 one year ago

bbnl1990 Group TitleBest ResponseYou've already chosen the best response.1
First, eliminate sin(x+y): fxfy=0 cos(x)cos(y)=0 cos(x)=cos(y) x=y Plug x=y to either fx or fy: fx=cos(y)sin(y+y)=0 cos(y)=sin(2y) cos(y)=cos(pi/22y) y=(pi/2)2y 3y=pi/2 y=pi/6 Since x=y, x=pi/6
 one year ago

jonnymiller Group TitleBest ResponseYou've already chosen the best response.0
Very clever!! That makes a lot of sense!! It's amazing how easy it looks once you've seen it done. I have been puzzling over that one for days! Thanks!
 one year ago

bbnl1990 Group TitleBest ResponseYou've already chosen the best response.1
Ur welcome! :)
 one year ago
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